# Asymptotic behavior of roots of an equation involving exponential and logarithm

### Prelude

This Post is a continuation of this Original Post. The original problem asked is:

How many solutions does the following equation have:

$$a^x = \log_a(x) \,,\quad a \in (0,1) \wedge x \in\mathbb{R}^+_0$$

And already has been answered (see Claude Leibovici's answer for details).

### Observations

I found this problem interesting and I numerically investigated it a bit deeper. This led me to another question. Figures below show that such roots exist and are reals:

And roots can be computed numerically for several values of $$a$$: My observations, so far, are:

1. All roots must lie in $$(0,1)$$ because $$\log_a(x) > 0 \, \forall x \in (0,1)$$ and $$\log_a(x) < 0$$ elsewhere, and $$a^x > 0 \,\forall x \in \mathbb{R}$$;
2. Solving this problem involves complex analysis (such as the use of Lambert W function) but the result will stay in the real domain;
3. A "divergence" point occurs at $$(e^{-e},e^{-1})$$;
4. Solving the original problem is equivalent to solve (base conversion and Lambert W properties):$$\ln(a) = \frac{W\left(x \ln(x)\right)}{x} = \frac{\ln(x)}{x} \Leftrightarrow a_k = \exp\left[\frac{W_k\left(x \ln(x)\right)}{x}\right]$$
5. Roots become triple when $$a < e^{-e}$$ (dashed black vertical line, as shown by Claude Leibovici in his answer)
6. Roots have asymptotic behavior, it can be checked in term of $$a(x)$$ for two branches:

• one root tends to unity as $$a\rightarrow 1$$: $$\lim\limits_{x\rightarrow 1} a = 1$$ (green curve rightmost);
• two roots tend to zero as $$a\rightarrow 0^+$$: $$\lim\limits_{x\rightarrow 0^+} a = 0$$ (green and orange curves leftmost).

### Questions:

My main questions are:

• How can I prove that one root tends to unity when base $$a\rightarrow 0^+$$ (blue curve leftmost)? By taking the limit of the proper branch.
• Is the point 4 correct? Investigating the solution using Wolfram Alpha it seems both Leibovici and my expressions are equivalent. But the first form suffer a huge float arithmetic error with numpy library. Anyway they can be plotted using the latter form: Side questions are:

• How is called the point where branches diverge?
• Do end of the branches also have a specific name?
• Can we say that roots are multiple at the "divergence" point? If so, in what sense are they multiple? Claude Leibovici: Roots are multiple in the sense that three first degrees of Taylor expansion vanishes at $$x=e^{-1}$$ with $$a=e^{-e}$$.
• Is the green branch a specific one because it behaves smoothly?

Considering the function $$f(x)=a^x-\frac{\log (x)}{\log (a)}$$ its derivatives are $$f^{(n)}(x)=a^x \log^n(a)+(-1)^n \frac{(n-1)!}{x^n\, \log(a)}$$ The first derivative cancels at two points given by $$x_1=\frac{W_0\left(\frac{1}{\log (a)}\right)}{\log (a)}\qquad \text{and}\qquad x_2=\frac{W_{-1}\left(\frac{1}{\log (a)}\right)}{\log (a)}$$ which, in the real domain, exist if $$\frac{1}{\log (a)}\geq -\frac 1 e$$ that is to say if $$a \leq e^{-e}$$. If this is the case, $$f(x_1)<0$$ and $$f(x_2)>0$$ which explains the three roots.
What is interesting is to look at what happens when $$a = e^{-e}$$. For this value, the solution of $$f(x)=0$$ is unique $$x=\frac 1e$$. At this point, the second derivative is also zero and the Taylor expansion is $$\frac{e^2}{6} \left(x-\frac{1}{e}\right)^3-\frac{5e^3}{24} \left(x-\frac{1}{e}\right)^4+O\left(\left(x-\frac{1}{e}\right)^5\right)$$ which makes that, at ths point, $$x=\frac 1e$$ is a triple root of the equation.
On another side, we could also solve the equation for $$a$$ and its solutions are given by $$a_1=\left(\frac{x \log (x)}{W_{0}(x \log (x))}\right)^{\frac{1}{x}}\qquad \text{and}\qquad a_2=\left(\frac{x \log (x)}{W_{-1}(x \log (x))}\right)^{\frac{1}{x}}$$ which do exist if $$x \leq \frac 1e$$. These two functions are worth to be plotted.
When $$x \to 1$$ the expansion of $$a_1$$ is $$a_1=1+(x-1)-(x-1)^2+\frac{1}{2} (x-1)^3+O\left((x-1)^4\right)$$ and using series reversion $$x= 1+(a_1-1)+O\left((a_1-1)^2\right)$$ making that if $$x\to 0 \implies a_1 \to 0$$.
• Thank you for taking time to answer. I have updated my OP to integrate information you provided. I had problem to plot you expression of $a$ using numpy but I succeed with another form (looks like a float error propagation I think it is due to the fractional power that grows quickly as $x \rightarrow 0$). Anyway Wolfram seems to show that both our versions agree. – jlandercy Dec 4 '18 at 9:17